R-matrix and transfer-matrix conventions

An R-matrix formula depends on tensor order, spectral normalization and whether the permutation has been included. This reference fixes those choices for the finite, homogeneous, periodic spin-½ XXX chain and derives the conversions most likely to change a sign or an energy zero. All tensor products here are ordinary, ungraded products of finite-dimensional spaces. Fermionic graded permutations require additional signs and are outside this convention set.

Helpful background. The spin-chain conventions define the physical Hamiltonian and active translation. The R-matrix lesson develops the three-factor algebra, and the Library proof establishes the commuting family.

Tensor legs and permutations

Let $V=\mathbb C^2$, with ordered basis $(|\uparrow\rangle,|\downarrow\rangle)$. On $V\otimes V$, define $P(v\otimes w)=w\otimes v$. In the ordered basis $(|\uparrow\uparrow\rangle,|\uparrow\downarrow\rangle,|\downarrow\uparrow\rangle,|\downarrow\downarrow\rangle)$,

$$P=\begin{pmatrix} 1&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&1 \end{pmatrix},\qquad P^2=I.$$

An index identifies a tensor factor, not a matrix entry. Thus $P_{13}|a,b,c\rangle=|c,b,a\rangle$, while $P_{12}|a,b,c\rangle=|b,a,c\rangle$. Products act from right to left. For a two-factor operator $X$,

$$X_{12}=X\otimes I,\qquad X_{23}=I\otimes X,\qquad X_{13}=P_{23}(X\otimes I)P_{23}.$$

The last formula embeds the operator without silently changing the order of its two input legs. If $X$ is not symmetric under interchange, $X_{21}=P_{12}X_{12}P_{12}$ need not equal $X_{12}$. For the rational matrix below they do agree.

Rational and checked R-matrices

Our dimensionless spectral parameter is $u\in\mathbb C$, and

$$R(u)=uI+iP, \qquad \check R(u)=P R(u)=uP+iI.$$

In particular $R(0)=iP$ and $\check R(0)=iI$. A checked R-matrix includes an interchange of the two factors. Keeping the same symbol while inserting or removing that interchange changes which Yang–Baxter relation is being written. With the present definitions,

$$\begin{aligned} R_{12}(a)R_{13}(a+b)R_{23}(b) &=R_{23}(b)R_{13}(a+b)R_{12}(a),\\ \check R_{12}(a)\check R_{23}(a+b)\check R_{12}(b) &=\check R_{23}(b)\check R_{12}(a+b)\check R_{23}(a). \end{aligned}$$

These are the ordinary and adjacent-leg forms. They hold for all complex $a,b$. They can be checked by using $P_{12}P_{13}=P_{13}P_{23}=P_{23}P_{12}$ and the reversed three-cycle identities; the lesson gives a direct expansion. The rational normalization and local exchange relation agree with Faddeev 1996, § 3, eqs. (35)–(39), PDF.

The symmetric and antisymmetric projectors,

$$\Pi_+=\frac{I+P}{2},\qquad \Pi_-=\frac{I-P}{2},$$

resolve the matrix as

$$R(u)=(u+i)\Pi_+ +(u-i)\Pi_-.$$

For spin ½, their ranks are three and one. Thus $R(i)$ has rank three, $R(-i)$ has rank one, and

$$R(u)^{-1}=\frac{uI-iP}{u^2+1},\qquad u\ne\pm i.$$

An exchange relation remains meaningful at $u=\pm i$, but a proof that multiplies by $R(u)^{-1}$ must first work away from those points. Transfer matrices in this polynomial normalization commute there as well, by continuation of the resulting polynomial identity.

Algebraic inversion and unitary matrices

Direct multiplication gives

$$R(u)R(-u)=-(u^2+1)I.$$

This scalar inversion identity is often called algebraic unitarity. It is not the statement $R(u)^\dagger R(u)=I$. For real $u$,

$$R(u)^\dagger R(u)=(u^2+1)I, \qquad \widehat R(u)=\frac{R(u)}{u+i}, \qquad \widehat R(u)^\dagger\widehat R(u)=I.$$

The normalized matrix has eigenvalues $1$ and $(u-i)/(u+i)$, each of modulus one for real $u$. For complex $u$, no such Hilbert-space unitarity follows. The scalar normalization preserves the Yang–Baxter relation wherever it is defined, but introduces a pole at $u=-i$; it does not repair the singular rank of $R(i)$.

Spectral shift and monodromy order

The physical Hilbert space is $\mathcal H=V_1\otimes\cdots\otimes V_N$, with $N\geq3$. Introduce a separate auxiliary copy $V_a$. Define

$$\begin{aligned} L_{an}(\lambda)&=R_{an}(\lambda-i/2) =(\lambda-i/2)I+iP_{an},\\ T_a(\lambda)&=L_{aN}(\lambda)\cdots L_{a1}(\lambda),\\ \tau(\lambda)&=\operatorname{tr}_aT_a(\lambda). \end{aligned}$$

$T_a$ acts on $V_a\otimes\mathcal H$; $\tau$ acts only on $\mathcal H$. The trace is over the auxiliary index, not over the spin-chain states. The spin form follows from $P_{an}=2\mathbf S_n\cdot\mathbf S_a+I/2$:

$$L_{an}(\lambda)=\lambda I+i\sum_{\alpha=x,y,z}S_n^\alpha\sigma_a^\alpha.$$

Thus the two common choices of spectral variable are related by $u=\lambda-i/2$. An unshifted monodromy built from $R_{an}(u)$ has transfer matrix $\mathcal T(u)=\tau(u+i/2)$. Its regular point is $u=0$, corresponding to $\lambda_0=i/2$ here. The local operator and ordered product are those of Faddeev 1996, § 3, eqs. (31), (35) and (42), PDF.

At the regular point each factor is $iP_{an}$. Following the auxiliary state through the ordered swaps and then taking its trace gives

$$\tau(\lambda_0)=i^N U,\qquad U|s_1,s_2,\ldots,s_N\rangle =|s_N,s_1,\ldots,s_{N-1}\rangle.$$

This active shift moves a spin flip from site $x$ to $x+1$. With $|k\rangle=N^{-1/2}\sum_xe^{ikx}|x\rangle$, it obeys $U|k\rangle=e^{-ik}|k\rangle$. Reversing the monodromy product to $L_{a1}\cdots L_{aN}$ instead gives $i^N U^{-1}$ at the regular point. Check the action on a basis state before comparing a source’s momentum sign.

Creation-block and coordinate rapidities

The same symbol $\lambda$ is often used for a coordinate rapidity and for the parameter of a monodromy creation block. With the product order above, they have opposite signs when they describe the same one-magnon wave. Write them as $\lambda_{\rm coord}$ and $\lambda_B$ while making this comparison.

In the ordered auxiliary basis $(\uparrow,\downarrow)$, write

$$T_a(\lambda_B)= \begin{pmatrix}A(\lambda_B)&B(\lambda_B)\\C(\lambda_B)&D(\lambda_B)\end{pmatrix}_a.$$

The upper-right block creates a down spin from $|0\rangle=|\uparrow\cdots\uparrow\rangle$. Define $\alpha=\lambda_B+i/2$ and $\beta=\lambda_B-i/2$. Before the auxiliary down spin flips at physical site $x$, each visited up spin contributes $\beta$; after that flip, each remaining up spin contributes $\alpha$. The flip contributes $i$. Therefore

$$B(\lambda_B)|0\rangle =i\sum_{x=1}^{N}\alpha^{N-x}\beta^{x-1}|x\rangle.$$

This follows directly from the local triangular action used in Faddeev 1996, § 4, equations (82)–(87), PDF. For $\alpha\beta\ne0$, neighboring coefficients have ratio

$$z_B=\frac{\beta}{\alpha} =\frac{\lambda_B-i/2}{\lambda_B+i/2}.$$

For real $\lambda_B$, set $z_B=e^{ik_{\rm coord}}$. If $z_B^N=1$, the vector is a periodic one-magnon eigenstate. Relabeling its coefficients under the active shift gives

$$U B(\lambda_B)|0\rangle =z_B^{-1}B(\lambda_B)|0\rangle =\frac{\lambda_B+i/2}{\lambda_B-i/2}B(\lambda_B)|0\rangle.$$

The ring condition is essential to the coefficient at the closing bond; a generic $B(\lambda_B)|0\rangle$ is not a translation eigenvector. The eigenphase agrees with Faddeev, § 4, equations (106)–(108), PDF. His momentum $p(\lambda_B)$ is defined by this phase $e^{ip}$; it is $-k_{\rm coord}$ modulo $2\pi$ in the site’s positive-exponent wave convention.

By contrast, the coordinate convention is

$$e^{ik_{\rm coord}} =\frac{\lambda_{\rm coord}+i/2}{\lambda_{\rm coord}-i/2}, \qquad \boxed{\lambda_{\rm coord}=-\lambda_B.}$$

For the direction-sensitive example $N=4$, $\lambda_B=1/2$ gives

$$\begin{gathered} z_B=-i,\qquad k_{\rm coord}=-\pi/2,\qquad \lambda_{\rm coord}=-1/2,\\ U B(1/2)|0\rangle=iB(1/2)|0\rangle, \qquad E=J(1-\cos k_{\rm coord})=J. \end{gathered}$$

In the ordered one-down-spin basis $(|1\rangle,|2\rangle,|3\rangle,|4\rangle)$, direct multiplication gives

$$B(1/2)|0\rangle =\frac14(-1-i,-1+i,1+i,1-i)^{\mathsf T}.$$

The state with $\lambda_B=-1/2$ instead has $k_{\rm coord}=\pi/2$, translation eigenvalue $-i$, and the same energy $J$. Energy alone therefore cannot detect this reversal. Neither can a total-momentum test using only zero or $\pi$, whose translation phases equal their inverses; the site’s symmetric root-pair examples need this additional check. Their established energies and explicitly verified singular vectors are unchanged.

This calculation proves the one-magnon conversion. It does not assert a full coordinate/algebraic wavefunction equivalence at arbitrary particle number. For regular two-root equations, simultaneous replacement $\lambda_j\mapsto-\lambda_j$ simply reciprocates both sides, while the energy depends on $\lambda_j^2$. Thus identical equation forms and energies cannot identify which wave convention a root labels. Singular parameters require the original polynomial vector and its limit, as in the singular-state reproduction.

Hamiltonian and scalar normalization

Write $D=\tau(\lambda_0)^{-1}\tau'(\lambda_0)$, with the derivative taken in the displayed spectral variable. Differentiating the product replaces one $iP$ by $I$. The traced terms give every periodic nearest-neighbor bond, including $(N,1)$:

$$D=-i\sum_{n=1}^{N}P_{n,n+1},\qquad P_{N,N+1}=P_{N,1}.$$

Consequently the site’s ferromagnetic, zero-vacuum-energy convention is

$$H=\frac J2\sum_{n=1}^{N}(I-P_{n,n+1}) =\frac{JN}{2}I-\frac{iJ}{2}D,\qquad J\gt0.$$

The source Hamiltonian has the opposite exchange sign: $H=-JH_{\mathrm{Faddeev}}$, using Faddeev 1996, § 3, eqs. (61)–(65), PDF. This conversion changes both the coefficient of the logarithmic derivative and the constant term. For $N=2$, the literal periodic sum counts the same physical pair twice; these ring examples use $N\geq3$.

Now multiply every local operator by a scalar $f(\lambda)$ analytic and nonzero near $\lambda_0$. Then

$$\widetilde\tau(\lambda)=f(\lambda)^N\tau(\lambda), \qquad \widetilde D=D+N\frac{f'(\lambda_0)}{f(\lambda_0)}I.$$

The same physical Hamiltonian must therefore be written

$$H=\left[\frac{JN}{2} +\frac{iJN}{2}\frac{f'(\lambda_0)}{f(\lambda_0)}\right]I -\frac{iJ}{2}\widetilde D.$$

For $f(\lambda)=1/(\lambda+i/2)$, the normalized Lax operator is $\widehat R(\lambda-i/2)$. Since $f'/f=i$ at $\lambda_0$, one obtains

$$\widetilde\tau(\lambda_0)=U,\qquad \widetilde D=D+iNI,\qquad H=-\frac{iJ}{2}\widetilde D.$$

The missing additive constant in this last formula has been absorbed by the scalar normalization. It is not a disagreement about the spin-chain spectrum. A further spectral reparameterization $\lambda=g(v)$ multiplies the logarithmic derivative by $g'(v_0)$, where $g(v_0)=\lambda_0$. Recovering the same Hamiltonian from that first derivative requires $g'(v_0)\ne0$ and the reciprocal change in its coefficient. If $g'(v_0)=0$, the first derivative vanishes and this extraction formula cannot be used.

What can be traced and differentiated

RTT relates two monodromies through an R-matrix acting only on their auxiliary spaces. That restriction makes the trace proof work. If $C$ acts only on an auxiliary factor, then $\operatorname{tr}_a(CX)=\operatorname{tr}_a(XC)$. It is false for arbitrary $C$ with physical operator entries. For example, on an auxiliary qubit and a physical qubit set

$$A=|0\rangle\langle0|\otimes\sigma_x, \qquad B=|0\rangle\langle0|\otimes\sigma_z.$$

Then $\operatorname{tr}_a(AB)=\sigma_x\sigma_z=-i\sigma_y$ whereas $\operatorname{tr}_a(BA)=i\sigma_y$. Treating either matrix as an auxiliary-only scalar matrix would invalidate the commutativity proof.

The logarithmic derivative $\tau^{-1}\tau'$ is defined only where $\tau$ is invertible. It is well-defined near $\lambda_0$ because $\tau(\lambda_0)=i^NU$. To generate higher coefficients, normalize $G(s)=\tau(\lambda_0)^{-1}\tau(\lambda_0+s)$ so $G(0)=I$, and use the local convergent series for $\log G(s)$. Here $G$ is a normalized transfer matrix, separate from the creation block $B$. A global matrix-logarithm branch is unnecessary. The charge lesson explains what the resulting commutativity proves and why independence, locality and spectral completeness require further arguments.

References

• Faddeev, Ludwig D. How Algebraic Bethe Ansatz works for integrable model. Les Houches lecture notes, arXiv:hep-th/9605187v1 (1996), §§ 3–4. Version record. Open PDF. The creation-block discussion uses equations (82)–(87) and (106)–(108).